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Suppose $X, Y$ are independent random variables with $X\sim NB(r, p)$ and $Y\sim NB(s, q)$. From a previous post, I understand that when the success probabilities are equal, $p = q$, then, $$ X+Y \sim NB(r + s, p). $$

What is the distribution of $X+Y$ in cases when $p \neq q$? If there is no analytical solution, I'd still be interested in an approximation.

RobPratt
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Will
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  • Are they still independent? Because then I'd imagine $P(X+Y = k) = \sum_i P(X = k - Y)P(Y = i)$ but you'd have to take into account the fact that $X$ can have at most $r$ successes and $Y$ can have at most $s$ successes. – FafaDog Mar 07 '23 at 22:35
  • Yes, they are still independent. I'm well beyond my depth here...grateful for any help doing this (no matter how pedantic/spelled out) – Will Mar 07 '23 at 22:43

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